# Single-slit diffraction intensity

1. Aug 1, 2012

### uzy5o

I have looked through my optics textbook and many websites about single-slit diffraction. They all end up deriving an equation that looks something like this: I = I0*(sinc(B))2, where B = (1/2)*k*b*sin(theta), k = wavenumber, b = slit width. I don't know if there's something I'm not understanding, but I have a hard time believing that the intensity only depends on the angle. Shouldn't intensity decrease as distance from the slit increases?
Thanks.

2. Aug 1, 2012

### Ibix

You are correct that the intensity does also vary with distance - the diffraction pattern is wider at greater distances so must be fainter. In the formula you quoted, I0 is the on-axis intensity (i.e. I(θ=0)=I0), and this is where the distance dependence has been "hidden". Generally, you don't care about distance dependence in far-field diffraction because the transverse distribution is where the interesting physics is, so your text has hidden the boring bits in the interests of clarity. Well spotted.

If you wanted to insert a distance term, the formula above tells you the way the pattern spreads perpendicular to the slit and you could measure laser beam width at different distances to get the spread parallel to the slit. The product of the two is the overall dependence of I0 on distance from the slit.

Does that make sense?

3. Aug 18, 2012

### alextx

Here is an animation of single slit diffraction for different widths of slits:
http://youtu.be/uPQMI2q_vPQ